Answer to Question #114247 in Physics for Derek Tye

Question

We do an experiment with a hoop (m=2.00 kg, r=10.0 cm) and an ideal spring. On a horizontal surface, we give the hoop an initial linear speed of v=1.75 m/s towards the horizontal spring, which is initially at equilibrium. The hoop rolls without slipping until it compresses the spring, coming to a momentary stop. Ignore frictional effects of the spring surface on the hoop. If the maximum distance the spring is compressed from its equilibrium is 6.40 cm, what is the spring constant k?

Answer 1

$$0.5mv^2+0.5I\omega^2=0.5kx^2$$

$$mv^2+mr^2\omega^2=kx^2\to 2mv^2=kx^2$$

$$2(2)(1.75)^2=(0.064)^2k$$

$$k=2990\frac{N}{m}$$